okay now let's talk about the mixed strategy nash equilibrium of uh the war effection game that we described well before this uh we need to uh sort of talk about the structure of the mixed strategy nash equilibrium because otherwise you know the different forms of mixed strategy nash uh will be required will require a different sort of calculation so here we are actually looking uh what's called stationary mixed strategy nash equilibrium well in in what sense it is stationary well at every period player i is going to drop out and wait uh with those probabilities with pi probability he's going to drop out with one minus pi probability he's going to keep waiting all right or keep fighting and this pi is going to be independent of the period so whether it's period 0 or period 10 or period 100 or 1000 they will be dropping out with those probabilities all right p1 for play 1 p2 for player 2. all right so under those stationary mixed strategy nash equilibrium i'm going to calculate things for player 1's perspective but because the game is symmetric i will just copy it down and change the ones with twos and get the strategy for player two so player ones expected utility when sorry so i'm gonna call them sigma 1 sigma 2 star all right which are basically you know p2 p2 oops 1 minus p2 star and then p1 1 minus p1 star alright so that's what a strategy profile is the probability of dropping out and staying waiting so what is probability for expected utility i'm sorry of player 1 given that his opponents play according to sigma 2 star okay i don't know that yet sigma 2 star but i know that it is p2 1 minus p2 well what is the expected so in the uh well the way we find mixed strategy ash equilibrium is remember you have to be indifferent between these two actions drop and wait when you drop i know the payoff because the game will be over but when you wait the game is not over yet so therefore expected utility of waiting makes no sense so therefore it is in fact equivalent to saying you're indifferent between dropping and waiting or you're indifferent between dropping today and waiting today and dropping tomorrow you see what i mean so therefore what i'm going to calculate is expected utility of player one if he drops in period t all right versus his expected utility if he drops at t plus one all right very good well obviously here uh all those payoffs are going to i mean this payoff and this path are going to be the same if player 2 happened to be dropping in some period earlier than t right i mean because the game will be over so assuming that the game is not yet over right i mean we do not make any payoff calculations for um uh terminal histories we should be working on um non-terminal history so what does that mean that means the game has not yet ended before time t so therefore i am looking some history which is in fact non-terminal and so the game is continuing and then player one is going to drop out at that period or he's going to wait and drop out next period all right so what are those expected payoffs well if you drop in period t well you are actually not going to be the winner you're going to lose and so you're going to suffer the cost as simple as this so this is l1 nothing but l1 t right this is uh what we calculated here all right what about i wait in period t and then drop out wow okay so here's the thing if i wait in period t the game may actually end in period t right because my opponent is going to drop with probability p2 and weight with probability 1 minus p p2 so therefore with p2 probability he may actually drop at period t why while i i i was waiting so if he drops in period t the game will be over all right and in this case this is the pair of i'm going to get h1 t right and then with 1 minus p2 probability my opponent will also wait in period t so i waited he waited so what happens in period t plus one i'm dropping well given that i'm dropping this period it really is irrelevant whether my opponent is dropping in that periods in period t plus one or not because i am going to lose anyway remember if both players drop at the same time both lose so therefore given that i drop it doesn't matter what my opponent does so for that reason if my opponent waits in period t in period t plus one i am dropping and hence getting l one t plus one remember i waited one more period to drop so therefore i accumulated a little bit more cost so that's it if you're looking for mixed strategy nash equilibrium this payoff and this payoff must be the same so the equivalence basically means uh dropping out at time t must give the player one exactly the same payoff as waiting uh in period t and then dropping in period t plus one all right so all we have to do is basically solve this equality well here let's open up h1t remember h1t is nothing but uh p2 multiplied by l1t plus delta tvi and so don't forget the pi multiplied by delta tv 1 plus 1 minus p2 well what is l1 t plus 1. so if you remember l1t is equal to uh this finite sum up to t minus 1. so therefore l i t plus 1 is going to be l 1 t itself minus i'm sorry we have extra cost uh delta to the power t c i okay so equals to l1t all right well again uh let's sort of open up this one minus p2 oh well we don't have to so what i do have is p2 times l1t plus p 1 minus p 2 times l 1 t so if i add them up i'll just l 1 t that will cancels out the one on the left hand side so these guys are just gone well what do i have i have zero on the left hand side on the right hand side i have p 2 delta to the power t v 1 plus okay so what i have is i want to leave everything with p 2 on the right hand side and basically send everything else to the left-hand side and then solve p2 because this is what i want to do i want to find the equilibrium p1 and p2 values so i'm going to have minus delta to the power tci so i send it to this side as positive delta to the power t c1 i'm sorry by the way this is c1 and on the right hand side i'm going to have minus times minus is plus p2 delta to the power t c1 well one thing delta to the power ts will cancel out on both sides so i have no delta whatsoever and when i take this into p2 parenthesis and divide both sides by c1 plus v1 what i will find is p2 is equal to c1 divided by v1 plus c1 that's it so i can now put star here because this is the equilibrium strategy of player 2. well it may sort of look weird because this is the strategy of player 2 but it has nothing to do with player 2's parameters like v2 c2 etc well that's perfectly normal because don't forget the idea of mixed strategy equilibrium a mixed strategy is a probability distribution over your actions that make your opponent indifferent all right so therefore making your opponent indifferent means your opponent's payoff relevant parameters will kick into uh the calculation so therefore the strategy of player two actually depends on what player one's uh payoff parameters are or i mean this is what i mean so well symmetrically obviously i don't need to calculate the entire thing once again for player uh one two's perspective this is player one's perspective and so we calculate p2 uh but i can just change the ones with two and uh i will get uh player two's i'm sorry player one's mixed strategy equilibrium so p1 star p2 star basically gives me the stationary mixed strategy nash equilibrium of this game that's it let's make some observations well first of all assuming that c2 is very very small all right both p1 star and p2 star are going to be very close to zero that means in equilibrium players are going to drop each period with a very small probability and with a very large probability almost close to one they will actually keep waiting or keep fighting so in a sense with a very big probability this game will actually continue for a very long period of time well what is the likelihood that this game is going to continue forever well this probability will approach to zero obviously because forever means like you multiply each of those one minus p1 pi's because they're independent sort of events you multiply 1 minus pi times 1 minus pi times 1 minus pi so infinitely many times so eventually uh for for large enough period this probability will converge to zero but the thing is with a very big probability this game will actually continue for a very long time uh well but the cost is going to get you know uh accumulated and so you know once the game is over it's very likely that these guys are actually going to suffer a negative payoff but the question is what is the expected payoff of playing those strategies i mean before these guys start playing this game how much payoff do they expect well i think that's simple because remember each player is indifferent between dropping out at time t and waiting and dropping out a period later so therefore and this t has nothing to do in our calculations as you see because we're looking at stationary mixed strategies equilibrium so player one's expected payoff is basically uh you know combination of these two payoffs i mean player one's expected payoff is combination of these two payouts but the thing is it doesn't matter because those two pairs are going to be equivalent right they're going to give us the same thing so the question is the not the question i'm sorry the expected path of player one is equal to his payoff of dropping immediately at time t equals zero you see what i mean so let me write this as this so the expected payoff of player i is equal to l i t equals zero okay once again the reasoning is that at every period the players are indifferent between dropping now or not dropping uh but this is true for any t and so therefore it's also true for t equals zero and so i am mixing between dropping now and not dropping uh given my opponent's you know mixture i should just multiply those numbers with my opponent's strategies and then calculate the expected payoff but i don't really need to do that because this number and this number are equivalent and so whatever my opponent's strategy is it's going to be probability times this plus probability times that and nevertheless it's just uh you know because these two numbers are the same it's just equal to one of them so therefore what is the payoff of dropping out immediately well it's zero right so therefore both player one and player two if they play one of those mixed strategies they both are going to get zero so in the mixed strategy nash equilibrium the outcome is zero zero in expected terms obviously what about the pure strategies in the pure strategy version of the equilibrium one player is going to get zero right for example here player one drops out and so he gets zero the other player however gets v2 this one player two drops out the symmetric player two gets zero payoff player one gets v1 payoff so when you look at or compare uh this is the the third uh or the second mixed strategy pure strategy nash equilibrium sorry when you compare this zero zero clearly it's inefficient right because it is possible that one of the players are going to make positive payoff um and so here actually nobody makes positive payoff but nevertheless uh this mixed strategy has very appealing and nice properties like it is for example evolutionary stable equilibrium which we haven't defined yet and i don't think we are going to cover it in this course um but but if you think of situations like war of attrition in real life actually uh playing a mixed strategy or the outcome of a mixed strategy makes a lot of sense well why is that so well because this is kind of a coordination problem as well